An Introduction to the Bergman Projection and Kernel

A Thesis

Presented in Partial Fulﬁllment of the Requirements for the Degree Master of Mathematical Sciences in the Graduate School of The Ohio State University

Philip Koch, B.S.

Graduate Program in the Department of Mathematics

The Ohio State University

2016

Master’s Examination Committee:

Dr. Kenneth Koenig, Advisor Dr. Rodica Costin Dr. Jeﬀery McNeal c Copyright by

Philip Koch

2016 Abstract

In this thesis, we discuss the fundamental concepts of the Bergman projection and kernel. In particular, we prove that the Bergman kernel on a given domain Ω can be characterized in terms of an orthonormal basis of A2(Ω). This result is central to the theory of the Bergman kernel and will therefore be shown in full detail. We calculate the Bergman kernel on several domains such as the unit disk, the unit ball in C2 and

Cn, and the annulus. In addition, we show that given two conformally equivalent domains Ω1, Ω2 ⊆ C, one can represent the Bergman kernel of Ω1 in terms of the

Bergman kernel of Ω2.

ii I dedicate this thesis to my wife, my parents, and my brothers.

iii Acknowledgments

I would like to thank several people who helped make this possible. First, I thank my advisor, Dr. Kenneth Koenig, for his support and guidance. His knowledge and wisdom were invaluable to my progress and growth as a mathematician and I feel extremely lucky to have worked with him.

Second, I would like to thank my committee members, Dr. Jeﬀery McNeal and Dr.

Rodica Costin. Their comments and insight were greatly appreciated. In particular,

Dr. Costin’s ﬁnal edits were very beneﬁcial.

Third, I thank my wife, Ayla, for her love and encouragement. She was my rock throughout this process. For taking care of the dishes and soda cans that I left on the table while working, for comforting me when I needed a break, and for all the other little things you do, thank you.

Fourth, I thank my brother, Thayer, for his advice and friendship. He was there when I needed someone to talk to and always available to play a video game when I wanted to relax.

Last I thank my parents, Lorin and Tandy, for their love and encouragement. So much of what I am today is because of them. You both are truly wonderful people and I could not have asked for better parents.

iv Vita

June 2012 ...... B.S. Mathematics, University of Washington 2014-present ...... Graduate Teaching Associate, Ohio State University.

Fields of Study

Major Field: Mathematics

v Table of Contents

Page

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita...... v

1. Motivation ...... 1

2. The Bergman Projection and Kernel ...... 9

3. The Bergman Kernel in C2 and Cn ...... 28

4. The Bergman Kernel and Conformal Mappings ...... 36

Bibliography ...... 47

vi Chapter 1: Motivation

We start by stating our topic of study: integral operators and their associated

kernels over C (and later we will look into those over Cn). In particular we begin with the Cauchy integral formula. A list of notation used in this thesis appears at

the end of this chapter.

Theorem 1.1. Let Ω ⊆ C be a bounded domain with piecewise smooth boundary. If f is holomorphic on Ω and f extends smoothly to the boundary of Ω, then

1 Z f(w) f(z) = dw (1.1) 2πi ∂Ω w − z for any point z ∈ Ω. [1, p. 113]

The Cauchy integral formula is an amazing fact and the very thing that piqued this author’s interest into complex analysis. We ask ourselves “what else can be said about this?” For one thing, it turns out for each f ∈ C(∂Ω), the Cauchy integral is holomorphic in Ω, which we’ll show in Corollary 1.3 below.

Theorem 1.2. Let F (z, s) be deﬁned for (z, s) ∈ Ω × [0, 1], where Ω is an open set in C. Suppose F satisﬁes the following properties:

(i) F (z, s) is holomorphic in z for each s.

(ii) F is continuous on Ω × [0, 1].

1 Then the function deﬁned on Ω by

Z 1 f(z) = F (z, s) ds 0 is holomorphic. [3, p. 56]

Corollary 1.3. Let Ω ⊆ C be a bounded open subset with smooth boundary and let f ∈ C(∂Ω). Then the Cauchy transform of f,

1 Z f(w) C(f)(z) = dw (1.2) 2πi ∂Ω w − z

is holomorphic in Ω.

Thus the Cauchy transform, C, can be thought of as an integral operator with 1 associated kernel C(z, w) = . 2πi(w − z)

Proof: Let γ : [0, 1] → ∂Ω be a smooth parameterization of ∂Ω and let z ∈ Ω. Then, by setting w = γ(t) in (1.2),

1 Z f(w) 1 Z 1 C(f)(z) = dw = F (z, t) dt, (1.3) 2πi ∂Ω w − z 2πi 0 f(γ(t)) where F (z, t) = γ0(t). We claim that F is holomorphic in z for each t and γ(t) − z that F is continuous on Ω × [0, 1] in order to apply Theorem 1.2.

Notice that for any t ∈ [0, 1], z 6= γ(t) for all z ∈ Ω. For if so, then there exists

◦ t0 ∈ [0, 1] such that z = γ(t0) ∈ ∂Ω. This would imply that z 6∈ Ω . Since Ω is open,

we would therefore have z 6∈ Ω. Hence we have a contradiction. And so, γ(t) − z 6= 0

for all t ∈ [0, 1] and z ∈ Ω. f(γ(t)) Therefore, for each t ∈ [0, 1], F (z, t) = γ0(t) is holomorphic in z. Further, γ(t) − z as f ∈ C(∂Ω), F (z, t) is continuous on Ω × [0, 1]. Thus, by Theorem 1.2, C(f)(z) is

holomorphic. 2 With all the wonderful results that complex analysis holds we have to ask ourselves several questions. Was the Cauchy integral formula a ﬂuke or are there more equations like this? If so, what properties do they possess? In addition, are they generalizable?

We plan to address each of these questions. In the coming chapters we will discuss the latter two, but for now we draw our attention to the ﬁrst. Before doing so we recall two classic results in complex analysis mentioned below.

Theorem 1.4. Let f be a holomorphic function on an open set Ω ⊆ C. Then for any zo ∈ Ω and 0 < r < dist(z0, ∂Ω)

1 Z 2π 1 Z f(z ) = f(z + reiθ) dθ = f(z) dV (z). (1.4) 0 2π 0 πr2 0 Dr(z0)

The ﬁrst equality of (1.4) is the mean value property for circles and the second is the mean value property for disks. We will utilize the latter in the proof of an important result in chapter 2. Whereas, for our purposes, the former will only be used in obtaining the property for the disk.

Proof: Let f be a holomorphic function on an open set Ω ⊆ C and let z0 ∈ Ω. Then for any 0 < r < dist(z0, ∂Ω), Dr(z0) ⊆ Ω. Hence, by the Cauchy integral formula,

1 Z f(w) f(z ) = dw. (1.5) 0 2πi w − z ∂Dr(z0) 0

iθ iθ By the substitution w = z0 + re for 0 ≤ θ ≤ 2π, dw = ire and

1 Z 2π f(z + reiθ) 1 Z 2π 0 iθ iθ (1.6) f(z0) = iθ ire dθ = f(z0 + re ) dθ 2πi 0 re 2π 0

3 iθ thereby establishing the ﬁrst equality. Next, by setting z = z0 + se for 0 ≤ s ≤ r and 0 ≤ θ ≤ 2π we have 1 Z 1 Z r Z 2π f(z) dV (z) = f(z + seiθ)s dθds 2π 2π 0 Dr(z0) 0 0 Z r = f(z0)s ds (1.7) 0 r2 = f(z ) . 0 2 Note that the middle equality was due to the mean value property for circles es- tablished in (1.6). Therefore, by dividing by r2/2, we obtain the second equality.

The answer to the first question is that “no, this wasn’t some fluke.” There are many different equations like the Cauchy integral formula. In particular, we will explore one in detail but do note that there are others to be studied as well.

Theorem 1.5 (The Bergman kernel on the unit disk). For all f ∈ A2(D) and z ∈ D, 1 Z f(w) f(z) = 2 dV (w). (1.8) π D (1 − zw) Notice that the integral deﬁned above is a convergent integral. Indeed, as

|z| < 1 and |w| < 1, |zw| = |z||w| < |z| and 1 − |z| > 0. Then, by the reverse triangle inequality, |1 − zw| ≥ 1 − |zw| > 1 − |z|. Thus, Z 2 Z 2 1 1 2 dV (w) ≤ 2 dV (w) < ∞, (1.9) D (1 − zw) D (1 − |z|) and we have the convergence of (1.8) by means of the Cauchy-Schwarz inequality.

Proof: Let f ∈ A2(D) and ﬁx z ∈ D. By setting w = reiθ with 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π, we have dV (w) = dxdy = rdrdθ. Hence,

Z f(w) Z 1 Z 2π f(reiθ) (1.10) 2 dV (w) = −iθ 2 r dθdr. D (1 − zw) 0 0 (1 − zre ) 4 −idw Since w = reiθ, dw = ireiθdθ = iwdθ. Therefore dθ = , e−iθ = r , and w w Z 1 Z 2π f(reiθ) Z 1 Z f(w) −idw r dθdr = r dr (1 − zre−iθ)2 r2 2 w 0 0 0 ∂D (1 − z w ) Z 1 Z −irf(w) dw = dr (1.11) w−zr2 2 w 0 ∂D ( w ) Z 1 Z −irwf(w) = 2 2 dwdr. 0 ∂D (w − zr ) . −irwf(w) Notice that as |z| < 1 and r ≤ 1, |zr2| < 1. Thus has a double pole at (w − zr2)2 zr2 ∈ D. By residue calculus, the inside integral of our last equality in (1.11) becomes Z −irwf(w) −irwf(w) 2 2 2 dw = 2πi · Res 2 2 , zr . (1.12) ∂D (w − zr ) (w − zr ) Since zr2 is a double pole, −irwf(w) d −irwf(w) Res , zr2 = lim (w − zr2)2 (w − zr2)2 w→zr2 dw (w − zr2)2 d = lim (−irwf(w)) w→zr2 dw (1.13) = lim (−irf(w) − irwf 0(w)) w→zr2 = −irf(zr2) − ir3zf 0(zr2). This calculation transforms (1.12) into Z −irwf(w) 2 3 0 2 2 2 dw = 2πi(−irf(zr ) − ir zf (zr )) ∂D (w − zr ) (1.14) = π(2rf(zr2) + 2zr3f 0(zr2)).

2 3 0 2 d 2 2 At this stage we notice that 2rf(zr ) + 2zr f (zr ) = dr (r f(zr )). Hence,

Z 1 Z Z 1 −irwf(w) 2 3 0 2 2 2 dwdr = π(2rf(zr ) + 2zr f (zr )) dr 0 ∂D (w − zr ) 0 Z 1 d (1.15) = π r2f(zr2) dr 0 dr = πf(z). 1 Z f(w) Therefore f(z) = 2 dV (z). π D (1 − zw) 5 1 The function B(z, w) = is called the Bergman kernel for the unit π(1 − zw)2 disk. We note that B(z, w) is a conjugate symmetric function which is holomorphic in z and anti-holomorphic in w. Further, the integral operator corresponding to B reproduces holomorphic functions. These facts turn out to be true in a generalized setting as well and we will discuss this in the next chapter.

The properties of the Bergman kernel are well known. However, in our opinion there does not seem to be a suitable reference which goes through the proofs of these properties in full detail. In particular, the proof of the result we present in

Theorem 2.16 will require several preliminary results that we intend to highlight. As such, it is our goal in Chapter 2 to carefully show and prove many of the results corresponding to the Bergman kernel.

The remainder of this chapter is dedicated to notation that will be used throughout.

Notation:

1. N = {0, 1,... }.

2. dV (z) denotes Lebesgue measure in Cn.

n n 3. R>0 = {r ∈ R | r > 0} and R>0 = {(r1, . . . , rn) ∈ R | ri > 0 for all 1 ≤ i ≤ n}.

4. h·, ·iH : H × H → C denotes the inner product on a given Hilbert space, H. If the space is clear from the context, we will omit the subscript and simply write

h·, ·i. Note that this inner product is linear in the ﬁrst argument and conjugate

linear in the second.

6 5. || · ||H : H → [0, ∞) denotes the norm on a given Hilbert space, H. If the space

is clear from the context, we will omit the subscript and simply write. || · ||.

6. Dr(a) = {z ∈ C | |z − a| < r} with a ∈ C and r ∈ R>0. For the unit disk

centered at the origin we use D.

ρ 7. Aσ = {z ∈ C | σ < |z| < ρ} with 0 < σ < ρ < ∞.

8. Bn = {z ∈ Cn | |z| < 1}

9. C(Ω) = {f :Ω → C | f is continuous}.

10. O(Ω) = {f :Ω → C | f is holomorphic}. Z 2 11. L2(Ω) = f :Ω → C |f(z)| dV (z) < ∞ . Ω

12. A2(Ω) = {f ∈ L2(Ω) | f ∈ O(Ω)} is the subspace of L2(Ω) of holomorphic

functions.

n n n 13. Dr (a) = Dr1 (a1) × · · · × Drn (an) with a ∈ C and r ∈ R>0.

n n 14. T (a, r) = ∂Dr1 (a1) × · · · × ∂Drn (an) with a ∈ C and r ∈ R>0.

Multi-index notation:

n For the n-tuple (α1, . . . , αn) ∈ N , we write

1. α = (α1, . . . , αn)

2. |α| = α1 + ... + αn

3. α! = α1! ··· αn!

n For z = (z1, . . . , zn), w = (w1, . . . , wn) ∈ C and n-tuples α, we write

7 α α1 αn 1. z = z1 ··· zn

2. hz, wi = z1w1 + ... + znwn

3. z − w = (z1 − w1) ··· (zn − wn)

8 Chapter 2: The Bergman Projection and Kernel

In this chapter we develop some of the theory of the Bergman projection and its

associated integral kernel in one complex variable. In particular, we intend to develop

the theory in an arbitrary domain and utilize these results to calculate the Bergman

kernel for the unit disk in a diﬀerent manner than in Theorem 1.5. The two complex

variable and the n-complex variable cases will be addressed in the next chapter. For

this chapter we will work solely with functions from A2(Ω) (or A2(D) depending on the situation).

Evidently A2(Ω) is a subspace of L2(Ω), which follows immediately from the lin-

earity of diﬀerentiation. Indeed, if f, g are holomorphic and α, β ∈ C, then αf + βg is also holomorphic. Further, αf + βg is square integrable since L2(Ω) is a Hilbert space. In fact, we will show below that A2(Ω) is a closed subspace of L2(Ω). In order to do so we will require the following estimate.

2 √1 1 Lemma 2.1. Let f ∈ A (Ω). Then for all z ∈ Ω, |f(z)| ≤ π · dist(z,Ωc) · ||f||L2(Ω).

2 c Proof: Let f ∈ A (Ω) and z0 ∈ Ω and set r = dist(z0, Ω ). Then Dr(z) ⊆ Ω and by the mean value property for disks in Theorem 1.4, Z 1 |f(z0)| = f(z) dV (z) πr2 Dr(z0) (2.1) 1 Z ≤ |f(z)| dV (z). πr2 Dr(z0) 9 By applying the Cauchy-Schwarz inequality to (2.1), we see

1 Z 1/2 Z 1/2 |f(z )| ≤ |f(z)|2 dV (z) |1|2 dV (z) 0 πr2 Dr(z0) Dr(z0) 1 Z 1/2 = |f(z)|2 dV (z) (πr2)1/2 (2.2) πr2 Dr(z0) 1 Z 1/2 = √ |f(z)|2 dV (z) . π r Dr(z0)

2 Note that since |f(z)| ≥ 0 and Dr(z0) ⊆ Ω,

1 Z 1/2 1 Z 1/2 √ |f(z)|2 dV (z) ≤ √ |f(z)|2 dV (z) π r π r Dr(z0) Ω (2.3) 1 = √ ||f|| 2 . π r L (Ω)

2 1 1 Hence, as f ∈ A (Ω) and z ∈ Ω were arbitrary, |f(z)| ≤ √ · · ||f|| 2 0 π dist(z, Ωc) L (Ω) 2 for every f ∈ A (Ω) and z ∈ Ω.

This estimate is crucial to the proof of Theorem 2.3 below. The following fact is a basic result in the theory of holomorphic functions.

Theorem 2.2. If {fn} is a sequence of holomorphic functions on Ω and fn → f uniformly on compact subsets of Ω, then f is holomorphic on Ω. [3, p. 53]

Theorem 2.3. Let Ω ⊆ C be an open set. Then A2(Ω) is a closed subspace of L2(Ω).

2 2 Proof: Let {fn} ⊆ A (Ω) such that fn → f for some f ∈ L (Ω). We wish to show

2 2 2 that f ∈ A (Ω). As fn is convergent in L (Ω), fn must also be Cauchy in L (Ω).

2 2 Further, as {fn} ⊆ A (Ω), fn −fm ∈ A (Ω) for all m, n ∈ N. Let E ⊆ Ω be a compact subset and observe that for all z ∈ E, 0 < dist(E, Ωc) ≤ dist(z, Ωc).

2 Let > 0. Since {fn} is Cauchy in L (Ω), we must have the existence of an N ∈ N

√ c such that ||fn − fm||L2(Ω) < ( π · dist(E, Ω )) for all n, m ≥ N. By Lemma 2.1, we

10 have that for all z ∈ E and n, m ≥ N, 1 1 |f (z) − f (z)| ≤ √ ||f − f || 2 n m π dist(z, Ωc) n m L (Ω) 1 1 (2.4) ≤ √ ||f − f || 2 π dist(E, Ωc) n m L (Ω) < .

Thus, fn → f uniformly on E. As E was arbitrary, we have by Theorem 2.2 that f

2 2 2 is holomorphic. Hence, f ∈ A (Ω) and A (Ω) must be a closed subspace of L (Ω).

Remark 1. It is important to note the following implication from the proof above:

2 convergence in A (Ω) implies pointwise convergence in C. That is, if {fn} is a se-

2 quence of functions in A (Ω) such that {fn} converges in the norm, then {fn} con-

verges pointwise as a function in C.

Since A2(Ω) is a closed subspace, we have L2(Ω) = A2(Ω) ⊕ A2(Ω)⊥ and there

exists an orthogonal projection onto A2(Ω). This is called the Bergman projection,

which we will denote by B. It will turn out that the integral operator deﬁned by

the Bergman kernel is equal to the Bergman projection, where an integral operator

is deﬁned as follows:

2 2 Deﬁnition 2.1. Suppose TK : L (Ω) → L (Ω) is a linear operator given by the

formula Z TK (f)(z) = K(z, w)f(w) dV (w) (2.5) Ω 2 for f ∈ L (Ω) and z ∈ Ω. We call TK an integral operator and K is its associated

(integral) kernel. [4, p. 187]

In order to prove that the two Bergman operators actually coincide we will ﬁrst

construct the Bergman kernel and establish several of its key properties. This will

require the following theorem.

11 Theorem 2.4 (The Riesz representation theorem). Let ` be a continuous linear functional on a Hilbert space H. Then there exists a unique g ∈ H such that

`(f) = hf, giH for all f ∈ H.

Moreover, ||`||H = ||g||H. [4, p. 182]

Lemma 2.5. Let Ω ⊆ C be an open subset. Then for each ﬁxed z ∈ Ω, there exists

2 2 a unique gz ∈ A (Ω) such that f(z) = hf, gzi for all f ∈ A (Ω).

2 Proof: Fix z ∈ Ω and consider the linear operator `z : A (Ω) → C deﬁned by

2 2 `z(f) = f(z) for all f ∈ A (Ω). Notice that for any f ∈ A (Ω),

|`z(f)| = |f(z)| ≤ Cz||f||L2(Ω) by Lemma 2.1 and so `z is therefore bounded. Thus

`z is a continuous linear functional. Hence, by Theorem 2.4, there exists a unique

2 2 gz ∈ A (Ω) such that `z(f) = hf, gzi for all f ∈ A (Ω). But `z(f) = f(z). Thus f(z) = hf, gzi.

Lemma 2.6. Set K(z, w) = gz(w), where gz is as deﬁned above. Then K is conjugate symmetric and is a reproducing kernel for A2(Ω). That is, for all f ∈ A2(Ω) and z ∈ Ω, Z f(z) = K(z, w)f(w) dV (w). (2.6) Ω

2 Proof: If K(z, w) = gz(w), then for each f ∈ A (Ω) we have from Lemma 2.5, Z Z f(z) = hf, gzi = f(w)gz(w) dV (w) = K(z, w)f(w) dV (w). (2.7) Ω Ω

Next, let w ∈ Ω be ﬁxed. We claim that K(w, ·) ∈ A2(Ω). To show this, notice that

2 K(w, ·) = (gw(·)) = gw(·) ∈ A (Ω). Hence, by Lemma 2.5,

K(w, z) = gw(z) = hgw, gzi = hgz, gwi = gz(w) = K(z, w)

12 and so K is conjugate symmetric.

Thus by virtue of the Riesz representation theorem, we have created an integral kernel on A2(Ω). We show next that this function is the only such function which satisﬁes the hypotheses of the following lemma.

Lemma 2.7. Assume K˜ (·, w) ∈ A2(Ω) for each w ∈ Ω, K˜ (z, w) is a reproducing kernel, and that K˜ is conjugate symmetric. Then K˜ = K.

Remark 2. For simplicity of notation and clarity we will sometimes denote K˜ (·, w) ˜ or K(·, w) by Kw(·) or Kw(·), respectively, for ﬁxed w ∈ Ω.

Proof: Suppose K˜ is as described above. Since K˜ (z, w) ∈ A2(Ω) for each ﬁxed w ∈ Ω, we see that by Lemma 2.5 ˜ ˜ ˜ K(z, w) = Kw(z) = hKw, gzi Z ˜ (2.8) = K(ζ, w)gz(ζ) dV (ζ). Ω

But gz(ζ) = K(z, ζ) by deﬁnition. So (2.8) becomes Z ˜ ˜ Kw(z) = K(ζ, w)K(z, ζ) dV (ζ) Ω Z (2.9) = K˜ (ζ, w)K(z, ζ) dV (ζ). Ω

By assumption K˜ is conjugate symmetric. Thus, as K˜ is an integral kernel and

K(z, ζ) is holomorphic in ζ,

Z ˜ ˜ Kw(z) = K(z, ζ)K(w, ζ) dV (ζ) Ω = (K(z, w)) (2.10)

= K(z, w) ˜ and K = K. 13 We now have the machinery necessary to show that the Bergman projection is

identiﬁed by the Bergman kernel. Indeed, the four properties of K (holomorphic in the

ﬁrst variable, a reproducing kernel, conjugate symmetry, and uniqueness) mentioned

above will prove suﬃcient.

2 2 Theorem 2.8. Let TK : L (Ω) → L (Ω) denote the integral operator with associated

2 2 kernel K, given by Lemma 2.6. Then TK = B, where B : L (Ω) → L (Ω) denotes the

Bergman projection.

Proof: To show equality it suﬃces to show that TK is an orthogonal projection

2 onto A (Ω). For if so, we have by uniqueness that TK = B. To begin we note

that by Lemma 2.6, K reproduces square integrable holomorphic functions. Thus, if

2 f ∈ A (Ω), then TK (f)(z) = f(z) for all z ∈ Ω. Next we show that TK (g) = 0 for all

g ∈ A2(Ω)⊥.

Let g ∈ A2(Ω)⊥ and ﬁx z ∈ Ω. Then Z TK (g)(z) = K(z, w)g(w) dV (w) Ω Z (2.11) = g(w)K(w, z) dV (w) Ω by the conjugate symmetry of K (as shown in Lemma 2.6). By deﬁnition of the inner

2 production on L (Ω) we see from above that TK (g)(z) = hg, Kzi. Since

2 Kz = K(·, z) ∈ A (Ω), we must have that TK (g)(z) = 0. Thus, as z ∈ Ω was

arbitrary, g = 0.

2 2 2 Now, let f ∈ L (Ω). Since A (Ω) is a closed subspace f = f1+f2 where f1 ∈ A (Ω)

2 ⊥ and f2 ∈ A (Ω) . Then, by the linearity of TK ,

TK (f) = TK (f1 + f2) = TK (f1) + TK (f2) = f1. (2.12)

14 2 2 As this f ∈ L (Ω) was arbitrary we see from (2.12) that Im(TK ) = A (Ω). We

additionally have that for all f ∈ L2(Ω),

TK (TK (f)) = TK (f1) = f1 = TK (f). (2.13)

2 Thus TK is a projection onto A (Ω) and the only thing left to verify is that TK is self-

∗ adjoint. However, this is immediate. Indeed, recall that the kernel of TK is K(w, z).

∗ However, by the conjugate symmetry of K, K(w, z) = K(z, w). Thus TK and TK

∗ are identiﬁed by the same kernel and we have TK = TK . Hence, TK is an orthogonal

2 projection onto A (Ω) and we have TK = B.

So far we have only described K abstractly. Now we will give an explicit way of calculating K given an orthonormal basis of A2(Ω) as in Theorem 2.16 below. In order to do so will require some results about sequences of functions which are uniformly bounded. The next four items are taken from [3, pp. 225–227] (though the notation will be changed in some cases).

∞ Deﬁnition 2.2. A sequence {En}n=1 of compact subsets of Ω ⊆ C is called an exhaustion if:

◦ (i) En ⊆ En+1 for all n = 1, 2,...

(ii) Any compact set E ⊆ Ω is contained in En for some n. In particular,

∞ [ Ω = En. n=1

∞ If such a sequence {En}n=1 exists we say that Ω has an exhaustion.

Lemma 2.9. Any open subset Ω ⊆ C has an exhaustion.

15 Remark 3. The proof amounts to examining two cases: whether Ω ⊆ C is bounded or unbounded. When Ω is bounded, set En = {z ∈ Ω | dist(z, ∂Ω) ≥ 1/n}. In the event that Ω is unbounded, set En = {z ∈ Ω | dist(z, ∂Ω) ≥ 1/n and |z| ≤ n}. However, this restriction to C is not necessary since the same argument holds in the case where

Ω ⊆ Cn. Therefore, any open set Ω ⊆ Cn has an exhaustion.

Deﬁnition 2.3. A family F of holomorphic functions on Ω is said to be normal if every sequence in F has a subsequence that converges uniformly on every compact subset K ⊆ Ω.

Theorem 2.10 (Montel’s theorem). Suppose F is a family of holomorphic functions on Ω ⊆ C that is uniformly bounded on compact subsets of Ω. Then:

(i) F is equicontinuous on every compact subset of Ω.

(ii) F is a normal family.

We include a few details of the proof as they will be relevant in a moment. For the full proof, see page 226 of [3]. Now, to prove equicontinuity requires a clever argument involving the Cauchy integral formula. Indeed if E ⊆ Ω is a compact subset, then there exists r > 0 so that for z, w ∈ E with |z − w| < r, Z 1 1 1 |f(z) − f(w)| = f(ζ) − dζ 2πi ζ − z ζ − w ∂D2r(w) (2.14) 1 2π(2r) ≤ C |z − w|, 2π 2r2 E where CE is the uniform bound of F on E. Therefore, for any > 0, choose δ > 0 r so that δ < min r, . Then (2.14) implies that |f(z) − f(w)| < whenever CE |z − w| < δ, thereby proving equicontinuity.

16 To prove F is a normal family requires two diagonalization arguments. During the

ﬁrst digaonalization argument, we choose a sequence {wi} which is dense in Ω. With

this dense subset, we inductively construct subsequences {fj,n}n∈N so that fj,n(wk)

converges for all k ≤ j. We extract the diagonal subsequence {gn} = {fn,n}n∈N, which will converge uniformly on a compact subset E of Ω (equicontinuity is essential to

prove this). To create the subsequence which converges uniformly on every compact

subset of Ω requires pairing this result and Lemma 2.9 with a somewhat similar

diagonalization argument.

Remark 4. As mentioned in Remark 3, every open Ω ⊆ Cn also has an exhaustion.

Therefore, proving that a family, F, of holomorphic functions deﬁned on Ω ⊆ Cn is normal only amounts to proving equicontinuity. To the point, this means that Montel’s

theorem is true in n-dimensions as well. As we will require this generalization for

Corollary 2.14, we present the argument to show equicontinuity below. To do so will

require the use of the Cauchy integral formula in n-dimensions.

Theorem 2.11 (The Cauchy integral formula for the polydisc). Let

n f(z) = f(z1, . . . , zn) be continuous on Ω ⊆ C and holomorphic with respect to each

n variable separately. Then for every closed polydisc Dr (a) ⊆ Ω, 1 Z f(ζ) f(z) = n dζ1 . . . dζn (2.15) (2πi) T (a,r) (ζ1 − z1) ... (ζn − zn)

n for all z ∈ Dr (a). [2, p. 7]

We must also prove a small algebraic result before moving forward with the n-

dimensional case of Montel’s theorem. As it will improve readability (at least in our

opinion), we will use the following notation: For ζ, z, w ∈ Cn, i ≤ j and k ≤ l, set

l wk ∆ j = (ζi − zi) ... (ζj − zj)(ζk − wk) ... (ζl − wl). (2.16) zi 17 l 0 wk w0 In the event that z or w does not occur, we write ∆z0 or ∆ j , respectively. In 0 zi l l wk wk addition, if i > j, then we set ∆ j = ∆z0 . For the following lemma, recall the zi 0 multi-index notation z − w = (z1 − w1) ··· (zn − wn) mentioned in Chapter 1 (item

3).

Lemma 2.12. Let ζ, z, w ∈ Cn. Then 1 1 − = ζ − z ζ − w

0 1 n−2 n−1 w0 w1 w1 w1 ∆zn (z1 − w1) + ∆zn (z2 − w2) + ... + ∆zn (zn−1 − wn−1) + ∆ 0 (zn − wn) 2 3 n z0 = n . w1 ∆ n z1 (2.17)

Proof: Let ζ, z, w ∈ Cn. Then n 0 w1 w0 ∆ 0 − ∆zn 1 1 1 1 z0 1 − = 0 − wn = wn ζ − z ζ − w w0 1 1 ∆ n ∆ 0 ∆zn z1 z0 1 n 1 1 0 w1 w1 w1 w0 ∆ 0 − ∆zn + ∆zn − ∆zn z0 2 2 1 = n (2.18) w1 ∆ n z1 1 n 0 0 w1 w2 w0 w0 ∆ 0 (∆ 0 − ∆zn ) + ∆zn (z1 − w1) z0 z0 2 2 = n . w1 ∆ n z1 Notice that n 0 n 2 2 0 w2 w0 w2 w2 w2 w0 ∆ 0 − ∆ n = ∆ 0 − ∆ n + ∆ n − ∆ n z z2 z z3 z3 z2 0 0 (2.19) 2 n 0 0 w2 w3 w0 w0 = ∆ 0 (∆ 0 − ∆zn ) + ∆zn (z2 − w2). z0 z0 3 3 Hence, (2.18) expands to

2 n 0 1 0 w1 w3 w0 w1 w0 ∆ 0 (∆ 0 − ∆zn ) + ∆zn (z2 − w2) + ∆zn (z1 − w1) z0 z0 3 3 2 n . (2.20) w1 ∆ n z1 n 0 wk w0 Inductively we iterate the process done in (2.19) to ∆ 0 − ∆zn for each k ≤ n. z0 k That is, for each k ≤ n,

n 0 n k k 0 wk w0 wk wk wk w0 ∆ 0 − ∆ n = ∆ 0 − ∆ n + ∆ n − ∆ n z zk z zk+1 zk+1 zk 0 0 (2.21) k wn 0 0 wk k+1 w0 w0 = ∆ 0 (∆ 0 − ∆zn ) + ∆zn (zk − wk). z0 z0 k+1 k+1 18 Upon doing so, we will obtain our desired result.

Corollary 2.13. If F is a family of holomorphic functions on Ω ⊆ Cn that is uniformly bounded on compact subsets of Ω, then F is equicontinuous on every compact

subset of Ω.

Proof: Let F ⊆ Ω be a compact subset and let f ∈ F. Choose r > 0 suﬃciently

n small so that Tn(z, 3r) ⊆ Ω for all z ∈ F (here 3r = (3r, . . . , 3r) ∈ R and

Tn(z, 3r) = ∂D3r(z1) × ... × ∂D3r(zn)). Let z, w ∈ F so that |z − w| < r and consider

Tn(w, 2r). Then |zi−wi| < r for 1 ≤ i ≤ n (since max{|zi−wi| | 1 ≤ i ≤ n} ≤ |z−w|).

Further, if ζ ∈ Tn(w, 2r), then |ζi − wi| = 2r and r ≤ |ζi − zi| ≤ 3r for any 1 ≤ i ≤ n.

Therefore by the Cauchy integral formula on the polydisc (Theorem 2.11), we have Z 1 1 1 |f(z) − f(w)| = n f(ζ) − dζ (2πi) Tn(w,2r) ζ − z ζ − w Z 1 1 1 ≤ n |f(ζ)| − dζ (2π) Tn(w,2r) ζ − z ζ − w 1 (3r)n−1|z − w | + ... + (3r)n−1|z − w | ≤ C 1 1 n n V (T (w, 2r)), (2π)n F (2r)nrn n (2.22)

where V (Tn(w, 2r)) denotes the volume of Tn(w, 2r) and CF is the uniform bound

of F on F . Note the last inequality was due to Lemma 2.12 and the fact that

|ζi − zi|, |ζi − wi| ≤ 3r for any 1 ≤ i ≤ n.

Since

|z1 − w1| + ... + |zn − wn| ≤ n(max{|zi − wi| | 1 ≤ i ≤ n}) ≤ n|z − w|, (2.23)

we have from (2.22) that

Thus we have the n-dimensional case of Montel’s theorem. As such, we obtain a very useful corollary.

n Corollary 2.14. Let Ω ⊆ C be an open subset and {fn} be a sequence of holomorphic functions converging pointwise to a function f :Ω → C. Suppose {fn} is uniformly bounded on every compact subset E ⊆ Ω. Then fn → f uniformly on compact subsets of Ω.

n Proof: Let Ω ⊆ C be an open set and {fn} be as deﬁned above. Further let E ⊆ Ω

be compact and {fnk } be a subsequence of {fn}. Since {fn} is uniformly bounded on

compact subsets, so must be {fnk }. Therefore, by Corollary 2.13 and our remark to

Theorem 2.10, {f } is a normal family. Hence, there exists a subsequence of {f 0 } nk nk of {f } such that f 0 → g uniformly on E for some g. However, as f → f pointwise nk nk n we must have that g = f by uniqueness. Thus f 0 → f uniformly on E. nk

Since {fnk } was an arbitrary subsequence, it follows that every subsequence of

{fn} has a sub-subsequence converging uniformly to f. Now, to show that fn → f uniformly on E, we suppose to the contrary. Then there exists 0 > 0 and z0 ∈ E such

that for every M ∈ N, there exists a nM ≥ N such that |fnM (z0) − f(z0)| ≥ 0. Then

{fnM } deﬁnes a subsequence of {fn} which does not converge to f. This would mean

that {fnM } could not have a subsequence converging uniformly to f, which leads us to a contradiction. Hence fn → f uniformly on E. As this E ⊆ Ω was arbitrary, fn → f uniformly on compact subsets of Ω.

20 2 Below we will show that, given an orthonormal basis {φn} ⊂ A (Ω), K can be ∞ X represented as the series φn(z)φn(w), which will converge uniformly on compact n=0 subsets of Ω × Ω. Considering this, we see the necessity for Corollary 2.14 (since

K depends on more than one complex variable). The pointwise convergence of the

series above will almost be immediate. However, to prove that the series is uniformly

bounded will require a diﬀerent expression of the inner product on L2(Ω).

Lemma 2.15. Let H be a Hilbert space. Then for all g ∈ H,

||g||H = sup {|hf, giH|} ||f||H=1 Proof: Let g ∈ H. If g = 0, the result is immediate. So suppose g ∈ H such that g 6= 0. Let f ∈ H such that ||f||H = 1. Then, by the Cauchy-Schwarz inequality,

|hf, gi|H ≤ ||f||H · ||g||H = ||g||H. Therefore, by taking the supremum over all f ∈ H

g such that ||f||H = 1, we see sup {|hf, giH|} ≤ ||g||H. Second, as g 6= 0, set f = . ||g||H ||f||H=1

Then ||f||H = 1 and

2 g ||g||H |hf, giH| = , g = = ||g||H. (2.25) ||g||H H ||g||H

Hence, sup {|hf, giH|} ≥ ||g||H and we have equality. ||f||H=1 The lemma above gives us an alternate characterization of the norm in any given

Hilbert space. This will prove useful when we consider the fact that |hf, gzi| = |f(z)|

for the gz as described in Lemma 2.5.

Now we are ﬁnally ready to characterize K with respect to an orthonormal basis.

2 Theorem 2.16. For any choice of orthonormal basis {φn} of A (Ω),

∞ X K(z, w) = φn(z)φn(w), (2.26) n=0 which converges uniformly and absolutely on compact subsets F ⊆ Ω × Ω.

21 Proof: Notice that as A2(Ω) is a closed subspace of L2(Ω) and L2(Ω) is a Hilbert space, A2(Ω) must also be a Hilbert space. As such, A2(Ω) admits at least one orthonormal basis. Let {φn} be such an orthonormal basis and ﬁx w ∈ Ω. Therefore, since K(·, w) ∈ A2(Ω),

∞ X K(·, w) = hK(·, w), φn(·)iφn(·). (2.27) n=0

However, notice that hK(·, w), φn(·)i = hφn(·),K(·, w)i = φn(w) by Lemma 2.7.

That is to say, ∞ X K(·, w) = φn(·)φn(w) (2.28) n=0 for ﬁxed w ∈ Ω and so K(·, w) converges in the norm. Recall that pointwise convergence is dominated by L2(Ω) convergence in A2(Ω). Therefore K(·, w) converges ∞ X pointwise to φn(·)φn(w). Thus, in accordance with Corollary 2.14, it suﬃces to n=0 show that our summation in (2.26) is uniformly bounded on compact subsets of Ω.

Next, we note that

∞ ∞ 2 X 2 X 2 ||K(·, w)||L2(Ω) = |hφn(·),K(·, w)i| = |φn(w)| . (2.29) n=0 n=0 Now, by Lemma 2.15 and the conjugate symmetry of K,

||K(·, w)||L2(Ω) = ||K(w, ·)||L2(Ω) = ||gw(·)||L2(Ω)

= sup {|hf, gwi|} (2.30) ||f||L2(Ω)=1 = sup {|f(w)|},

||f||L2(Ω)=1 where the last equality is due to Lemma 2.5. Therefore, by Lemma 2.1 we must have

||K(·, w)||L2(Ω) = sup {|f(w)|} ||f||L2(Ω)=1 1 1 ≤ sup √ · · ||f|| 2 π dist(w, Ωc) L (Ω) (2.31) ||f||L2(Ω)=1 1 1 = √ · . π dist(w, Ωc)

22 Thus, if E is a compact subset of Ω, then there exists a constant CE ≥ 0 so that

||K(·, w)||L2(Ω) ≤ CE for all w ∈ E. From (2.29) this implies that

P∞ 2 2 n=0 |φn(w)| ≤ CE. Therefore, on compact subsets F ⊆ Ω × Ω, we have by the Cauchy-Schwarz inequality, that

∞ ∞ !1/2 ∞ !1/2

X X 2 X 2 2 φn(z)φn(w) ≤ |φn(z)| |φn(w)| ≤ CF . (2.32) n=0 n=0 n=0 P∞ That is, n=0 φn(z)φn(w) is uniformly bounded on compact subsets of Ω × Ω. Note that we may not apply Corollary 2.14 since this series is not holomorphic on Ω × Ω.

However if we make the following observation, we will obtain our result.

∗ ∗ Let Ω = {w | w ∈ Ω} and for each m ∈ N deﬁne fm :Ω × Ω → C by

m X fm(z, w) = φn(z)φn(w) (2.33) n=0

Then {fm} is a sequence of holomorphic functions that is uniformly bounded and

converges pointwise to K(z, w) by the arguments presented above. Therefore, by

0 ∗ Corollary 2.14, {fm} converges uniformly to K on compact subsets F ⊆ Ω × Ω .

However, for all (z, w) ∈ Ω × Ω and for each m ∈ N,

m X φn(z)φn(w) = fm(z, w), (2.34) n=0 P∞ which converges uniformly to K(z, w) = K(z, w). Hence, n=0 φn(z)φn(w) converges

uniformly to K(z, w) on compact subsets F ⊆ Ω × Ω.

Thus (2.26) provides a formula for the integral kernel on any bounded open set

Ω ⊆ C. However, we note that in general this can prove quite cumbersome to write explicitly depending on the domain Ω we reside in. We do not go into details in this chapter but even for “nice” domains, such as an annulus, or an ellipse centered at the origin, the Bergman kernel turns out to be diﬃcult to calculate.

23 For most domains, it is not possible to determine an explicit orthonormal basis.

Even when we can ﬁnd an orthonormal basis, it is usually diﬃcult to say whether

or not the summation deﬁned in (2.26) will simplify to a simple equation. However,

when Ω = D the result does turn out to be simple as we saw in Theorem 1.5. We will now calculate the Bergman kernel by means of Theorem 2.16 above, which will

require ﬁnding an orthonormal basis for A2(D).

Theorem 2.17. Let H be a Hilbert space. The following properties of an orthonormal

set {φn} are equivalent:

(i) Finite linear combinations of elements in {φn} are dense in H.

(ii.) If f ∈ H and hf, φni = 0 for all n, then f = 0.

PN (iii.) If f ∈ H and SN (f) = n=0 anφn, where an = hf, φni, then SN (f) → f as N → ∞ in the norm.

2 P∞ 2 (iv.) If an = hf, φni, then ||f|| = n=0 |an| . [4, p. 165]

q n+1 n Lemma 2.18. Let φn(z) = π z for all n ≥ 0. Then {φn} is an orthonormal

basis for A2(D).

For this proof if will be important to recall that for m, n ∈ Z, ( Z 2π 0 m 6= n ei(m−n)θdθ = (2.35) 0 2π m = n.

Proof: We begin by showing the orthonormality of the set {φn}. For distinct

m, n ∈ N, Z p ! (n + 1)(m + 1) n m hφn, φmi = z z dV (z). (2.36) D π

24 Setting z = reiθ for 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π, (2.36) becomes

Z 1 Z 2π p ! (n + 1)(m + 1) n inθ m −imθ hφn, φmi = r e r e rdθdr 0 0 π ! (2.37) Z 1 Z 2π p(n + 1)(m + 1) = rn+m+1ei(n−m)θ dθdr. 0 0 π

With (2.35) in mind, we see that hφn, φmi = 0 if n 6= m. However, if n = m, then

(2.36) becomes Z 1 Z 2π n + 1 2n+1 hφn, φni = r dθdr 0 0 π Z 1 (2.38) = 2(n + 1)r2n+1 dr 0 = 1.

Thus {φn} forms an orthonormal set and it remains to show that this set forms a

basis for A2(D).

2 2 Let f ∈ A (D) and suppose hf, φni = 0 for all n. Note that as f ∈ A (D),

P∞ k f ∈ O(D) as well. Thus, f has a unique power series representation, f = k=0 akz

with ak ∈ C for all k. Fix n ∈ N. Then, by converting to polar, Z 0 = hf, φni = f(z)φn(z) dV (z) ZD1 Z 2π iθ iθ = f(re )φn(re )r dθdr 0 0 ∞ ! r ! Z 1 Z 2π X n + 1 (2.39) = a rkeikθ rne−inθ r dθdr k π 0 0 k=0 r ∞ n + 1 Z 1 X Z 2π = a rn+k+1 ei(k−n)θ dθdr. π k 0 k=0 0 However, as seen in (2.35), all terms of the summation will vanish save for the nth term of the power series. So, continuing with the calculation,

rn + 1 Z 1 0 = hf, φ i = 2πa r2n+1 dr n π n 0 (2.40) rn + 1 π =a n π n + 1 25 So by necessity, an = 0. As n ∈ N was arbitrary, it follows from (2.40) that an = 0

for all n. That is, f = 0. Thus by Theorem 2.17, {φn} forms an orthonormal basis

2 for A (D).

If we may make a small remark, the calculations done in the proof above provide

us with a suﬃcient condition to determine whether a given holomorphic function on

D will be square integrable.

∞ 2 X |an| Theorem 2.19. Let f ∈ O( ). Then f ∈ A2( ) if the series is convergent, D D n + 1 n=0 where an represent the coeﬃcients of the power series representation of f.

∞ X n Proof: Let f ∈ O(D). The f has a power series representation, f = anz , which n=0 converges uniformly on compact subsets of D. In order for f ∈ A2(D), we require Z |f(z)|2 dV (z) < ∞. Then, by converting to polar, D Z Z |f(z)|2 dV (z) = f(z)f(z) dV (z) D D Z 1 Z 2π ∞ ! ∞ ! (2.41) X n inθ X m −imθ = anr e amr e r dθdr. 0 0 n=0 m=0 Recalling (2.35) we see that all terms of the (2.41) vanish save for when n = m.

Hence, Z 1 Z 2π ∞ ! ∞ ! Z 1 Z 2π ∞ X n inθ X m −imθ X 2 2n+1 anr e amr e dV (z) = |an| r dθdr 0 0 n=0 m=0 0 0 n=0 Z 1 ∞ X 2 2n+1 = 2π |an| r dr. 0 n=0 (2.42)

2 Since |an| ≥ 0 for all n ∈ N, we have by Tonelli’s theorem that Z 1 ∞ ∞ Z 1 X 2 2n+1 X 2 2n+1 2π |an| r dr = 2π |an| r dr 0 0 n=0 n=0 (2.43) ∞ 2 X |an| = π . n + 1 n=0 26 Z ∞ 2 X |an| That is, |f(z)|2 dV (z) = π . Thus, if f ∈ O( ), then it is suﬃcient to n + 1 D D n=0 ∞ 2 X |an| require that < ∞ in order for f ∈ A2( ). n + 1 D n=0 1 Continuing on, we recall that for |z| < 1, the function f(z) = is holomorphic 1 − z and has power series representation

∞ X 1 zn = . (2.44) 1 − z n=0

Then we may diﬀerentiate this series term by term to obtain the following lemma.

∞ X 1 Lemma 2.20. If |z| < 1, then (n + 1)zn = . (1 − z)2 n=0

We are now ready to calculate the Bergman kernel on D by means of Theorem 2.16. Indeed, if we use the orthonormal basis found in Lemma 2.18 we see that

∞ ∞ r ! r ! X X n + 1 n + 1 B(z, w) = φ (z)φ (w) = zn wn n n π π n=0 n=0 ∞ (2.45) 1 X = (n + 1)(zw)n. π n=0 Therefore, by Lemma 2.20, we have precisely that

1 B(z, w) = . (2.46) π(1 − zw)2

Notice that in our construction of the Bergman kernel, there was nothing that

required Ω to be speciﬁcally a subset of C. In fact, our argument is generalizable for

Ω ⊆ Cn. Thus, as alluded to in the beginning of the chapter we will now calculate the Bergman kernel on the unit ball for the two complex variable and n-complex variable

case.

27 Chapter 3: The Bergman Kernel in C2 and Cn

In this chapter we will construct the Bergman kernel on A2(B2) and then on

A2(Bn). In each case we will create an orthonormal basis and utilize Theorem 2.16.

n m For the two dimensional case, we note that {z1 z2 }m,n∈N form an orthogonal basis

2 2 on A (B ), but it is not orthonormal. Thus, we must ﬁnd coeﬃcients cmn ∈ C so

n m that {cmnz1 z2 }m,n∈N forms an orthonormal basis, which we will require calculating

m n ||z1 z2 ||L2(B2). Before doing so, we recall the Beta function which is deﬁned by:

Z 1 Γ(x)Γ(y) β(x, y) = tx−1(1 − t)y−1 dt = , (3.1) 0 Γ(x + y) where Γ denotes the Gamma function and x, y > 0. We will utilize this function on several occasions in calculating the integrals below, especially when we investigate the n-dimensional case.

Let m, n ∈ N. Then, Z m n 2 2m 2n ||z z || 2 2 = |z | |z | dV (z )dV (z ) 1 2 L (B ) 1 2 1 2 B2 Z Z ! (3.2) 2m 2n = |z1| √ |z2| dV (z2) dV (z1). 2 |z1|≤1 |z2|≤ 1−|z1|

28 iθ2 p 2 First set z2 = r2e for 0 ≤ r2 ≤ 1 − |z1| and 0 ≤ θ2 ≤ 2π. Then the inner integral

becomes √ 2 Z Z 2π Z 1−|z1| 2n 2n √ |z2| dV (z2) = r2 r2dr2dθ2 |z |≤ 1−|z |2 0 0 2 1 √ 2 Z 1−|z1| 2n+1 = 2π r dr2 2 (3.3) 0 √ 2 1−|z1| 2π 2n+2 = r2 2n + 2 0 π = 1 − |z |2n+1 . n + 1 1 Thus, the equation in (3.2) becomes Z m n 2 π 2m 2n+1 ||z z || 2 2 = |z | 1 − |z | dV (z ). 1 2 L (B ) 1 1 1 (3.4) n + 1 |z1|≤1

iθ1 Next by setting z1 = r1e with 0 ≤ r1 ≤ 1 and 0 ≤ θ1 ≤ 2π, Z Z 2π Z 1 2m 2n+1 2 m 2 n+1 |z1| 1 − |z1| dV (z1) = (r1) (1 − r1) r1dr1dθ1 |z1|≤1 0 0 Z 2π Z 1 1 m n+1 = t (1 − t) dtdθ1 (3.5) 2 0 0 Z 1 = π tm(1 − t)n+1 dt. 0 Note that the second equality above came from making a change of variables with

2 t = r1 (and hence, dt = 2r1dr1 and 0 ≤ t ≤ 1). Therefore,

Z 1 Γ(m + 1)Γ(n + 2) tm(1 − t)n+1 dt = β(m + 1, n + 2) = . (3.6) 0 Γ(m + n + 3)

Thus, by combining (3.4), (3.5), and the result directly above, we have m n 2 π Γ(m + 1)Γ(n + 2) ||z z || 2 2 = π . (3.7) 1 2 L (B ) n + 1 Γ(m + n + 3)

Recalling that for all n ∈ N, Γ(n + 1) = n!, we have m n 2 π m!(n + 1)! ||z z || 2 2 = π 1 2 L (B ) n + 1 (m + n + 2)! (3.8) π2m!n! = . (m + n + 2)!

29 Thus we have found our constants cmn and our corresponding orthonormal basis

{φmn(z1, z2)}m,n∈N, where for each m, n ∈ N, r (m + n + 2)! φ (z , z ) = zmzn. (3.9) mn 1 2 π2m!n! 1 2

Now, before we move on we recall Lemma 2.20 and realize that this formula can actually be generalized. The proof follows immediately from induction on k and we therefore omit it.

∞ X k! Lemma 3.1. If |z| < 1, then (n + 1) ··· (n + k)zn = (1 − z)k+1 n=0 This lemma will prove useful in our calculations of the Bergman kernel in both

the two dimensional case and the n-dimensional case.

Theorem 3.2. The Bergman kernel on B2 is given by the equation

2 B(z, w) = B(z1, z2, w1, w2) = (3.10) 2 3 π (1 − (z1w1 + z2w2))

2 2 for z = (z1, z2), w = (w1, w2) ∈ B ⊆ C .

2 2 Proof: Let φm,n ∈ A (B ) be as deﬁned in (3.9). Then {φmn}m,n∈N forms an or-

thonormal basis for A2(B2). Therefore, by Theorem 2.16, X B(z, w) = φmn(z1, z2)φmn(w1, w2) m,n∈N X (m + n + 2)! = (zmzn)(w mw n) π2m!n! 1 2 1 2 (3.11) m,n∈N X (m + n + 2)! = (z w )m(z w )n. π2m!n! 1 1 2 2 m,n∈N Now, we may rewrite (3.11) by indexing as follows:

∞ X X (m + n + 2)! m n B(z, w) = (z1w1) (z2w2) . (3.12) π2m!n! k=0 m+n=k 30 We note that we may represent n in terms of k and m, which transforms (3.12) to

∞ k X X (k + 2)! B(z, w) = (z w )m(z w )k−m π2m!(k − m)! 1 1 2 2 k=0 m=0 (3.13) ∞ k ! X (k + 2)(k + 1) X k! = (z w )m(z w )k−m , π2 m!(k − m)! 1 1 2 2 k=0 m=0 with the last equality due to the inner sum depending on m. But the inside summation

k is precisely the binomial expansion of (z1w1 + z2w2) . Hence, by Lemma 3.1 and this observation, ∞ X (k + 2)(k + 1) B(z, w) = (z w + z w )k π2 1 1 2 2 k=0 (3.14) 2 = 2 3 . π (1 − (z1w1 + z2w2)) and we have our desired result.

It is natural and pleasantly surprising to see some similarity between the Bergman kernel on D and on B2. In fact, this will continue on to Bn as well. However, notice that this notation will only become more complex (pun intended!) as we move to the n-dimensional case. Indeed, B(z1, . . . , zn, w1, . . . , wn) becomes painful to read and write! As such, we will use the multi-index notation as mentioned in Chapter 1 when necessary.

Using multi-index notation, the orthonormal basis in (3.9) can be written as

{φα}α∈N2 , where r (|α| + 2)! φ (z) = zα. (3.15) α π2α!

This can of course be extended to Nn for any n, and we do so below.

α Much like in the two complex variable case, {z }α∈Nn will form an orthogonal

2 n basis on A (B ). So, once again we ﬁnd cα to normalize this basis. We will proceed

α in the same fashion as before - by computing ||z ||L2(Bn). When relevant we shall use

31 n the following notation: for k > 1, z = (z1, . . . , zn) ∈ B we denote

p 2 2 Ak = 1 − |z1| − ... − |zk−1| . Further, notice that

2 2 2 Ak = Ak−1 − |zk−1| for all k > 2. (3.16)

For α ∈ Nn we have Z α 2 2α ||z || 2 n = |z| dV (z) L (B ) n ZB Z Z 2α1 2αn = |z1| ··· |zn| dV (zn) ··· dV (z1) |z1|≤1 |z2|≤A2 |zn|≤An Z Z Z 2π Z An 2α1 2αn+1 = |z1| ··· rn drndθndV (zn−1) ··· dV (z1), |z1|≤1 |z2|≤A2 0 0 (3.17)

iθn where the conversion to polar zn = rne with 0 ≤ rn ≤ An and 0 ≤ θn ≤ 2π was made in the last step. Hence, the inner integral (with respect to zn−1) in (3.17) becomes

π Z 2αn−1 2 2 αn+1 |zn−1| (An−1 − |zn−1| ) dV (zn−1), (3.18) αn + 1 |zn−1|≤An−1

iθn−1 which by setting zn−1 = rn−1e with 0 ≤ rn−1 ≤ An−1 and 0 ≤ θn−1 ≤ 2π, will yield Z 2π Z An−1 π 2αn−1+1 2 2 αn+1 rn−1 (An−1 − rn−1) drn−1dθn−1. (3.19) αn + 1 0 0

Now we make a substitution again. This time we set rn−1 = An−1u with 0 ≤ u ≤ 1.

Then drn−1 = An−1du and (3.19) becomes

2π2 Z 1 2αn−1+1 2αn−1+1 2 2 2 αn+1 An−1 u (An−1 − An−1u ) An−1 du, (3.20) αn + 1 0 which simpliﬁes to

2π2 Z 1 2(αn+αn−1+2) 2 αn−1 2 αn+1 An−1 (u ) (1 − u ) udu. (3.21) αn + 1 0

32 If we set t = u2 and recall the Beta function in (3.1), we are left with

π2 Z 1 π2A2(αn+αn−1+2) 2(αn+αn−1+2) αn−1 αn+1 n−1 An−1 t (1 − t) dt = β(αn−1 + 1, αn + 2) αn + 1 0 αn + 1 π2A2(αn+αn−1+2) Γ(α + 1)Γ(α + 2) = n−1 n−1 n αn + 1 Γ(αn + αn−1 + 3) 2 π αn!αn−1! 2(αn+αn−1+2) = An−1 (αn + αn−1 + 2)!

2(αn+αn−1+2) = Cn−1An−1 , (3.22) 2 π αn!αn−1! where Cn−1 = . (αn + αn−1 + 2)! Going back to (3.17), we have reduced our integral to Z Z Z 2α1 2αn−2 2(αn+αn−1+2) Cn−1 |z1| ··· |zn−2| An−1 dV (zn−2) ··· dV (z1). |z1|≤1 |z2|≤A2 |zn−2|≤An−2 (3.23)

2 2 2 If we recall (3.16), then we have An−1 = An−2 − |zn−2| . Thus, the inner integral of (3.23) becomes Z 2αn−2 2 2 αn+αn−1+2 |zn−2| (An−2 − |zn−2| ) dV (zn−2) (3.24) |zn−2|≤An−2

iθn−2 and now we see a pattern forming. Indeed, the conversion to polar zn−2 = rn−2e with 0 ≤ rn−2 ≤ An−2 and 0 ≤ θn−2 ≤ 2π converts (3.24) to

Z 2π Z An−2 2αn−2+1 2 2 αn+αn−1+2 rn−2 (An−2 − rn−2) drn−2dθn−2. (3.25) 0 0

Set rn−2 = An−2u with 0 ≤ u ≤ 1. Then drn−2 = An−2du and (3.25) is equivalent to Z 1 2αn−2+1 2αn−2+1 2 2 2 αn+αn−1+2 2π An−2 u (An−2 − An−2u ) An−2du, (3.26) 0 which simpliﬁes to

Z 1 2(αn+αn−1+αn−2+3) 2 αn−2 2 αn+αn−1+2 2π An−2 (u ) (1 − u ) udu. (3.27) 0 33 By setting t = u2 and utilizing the Beta function again, this integral is equivalent to

2(αn+αn−1+αn−2+3) πAn−2 β(αn−2 + 1, αn + αn−1 + 3) = (3.28) 2(αn+αn−1+αn−2+3) αn−2!(αn + αn−1 + 2)! = πAn−2 . (αn + αn−1 + αn−2 + 3)! Thus, we have reduced (3.23) to Z Z Z 2α1 2αn−3 2 2 αn+αn−1+αn−2+3 Cn−2 |z1| ··· |zn−3| (An−3 − |zn−3| ) |z1|≤1 |z2|≤A2 |zn−3|≤An−3

dV (zn−3) ··· dV (z1), (3.29) 3 π αn!αn−1!αn−2! where Cn−2 = . Iterating this process, we continue until we (αn + αn−1 + αn−2 + 3)! reach

πn−1α ! ··· α ! Z n 2 2α1 2 αn+···+α2+(n−1) |z1| (1 − |z1| ) dV (z1). (3.30) (αn + ··· + α2 + (n − 1))! |z1|≤1

iθ1 Then, by doing one last conversion to polar with z1 = r1e , we are left with

πn−1α ! ··· α ! Z 2π Z 1 n 2 2α1 2 αn+···+α2+(n−1) r1 (1 − r1) rdr1dθ1. (3.31) (αn + ··· + α2 + (n − 1))! 0 0

Last we set t = r2 with 0 ≤ t ≤ 1 and dt = 2rdr, which takes our double integral to Z 1 α1 αn+···+α2+(n−1) π t (1 − t) dt = πβ(α1 + 1, αn + ··· + α2 + n) 0 πΓ(α + 1)Γ(α + ··· + α + n) = 1 n 2 (3.32) Γ(αn + ··· + α1 + (n + 1)) πα !(α + ··· α + (n − 1))! = 1 n 2 . (αn + ··· + α1 + n)! Hence, in accordance with our multi-index notation,

n α 2 π α! ||z || 2 n = . (3.33) L (B ) (|α| + n)!

n Thus, we have found our constants cα and our orthonormal basis for B is

{φα}α∈Nn , where r (|α| + n)! φ = zα (3.34) α πnα! 34 and we are ﬁnally ready to calculate the Bergman kernel in the n-dimensional case.

Before we begin we recall the multinomial theorem.

Theorem 3.3. For any positive integer m and nonnegative integer n, n X n α (z1 + ··· + zm) = z , (3.35) α1, . . . , αm |α|=n n n! where = . α1, . . . , αn α!

Theorem 3.4. The Bergman kernel on Bn is given by the equation n! B(z, w) = (3.36) πn(1 − hz, wi)n+1 for z, w ∈ Bn ⊆ Cn.

2 n Proof: Let φα ∈ A (B ) be as deﬁned in (3.34). Then {φα}α∈Nn forms an orthonormal basis for A2(Bn). Hence, by Theorem 2.16, for z, w ∈ Bn X B(z, w) = φα(z)φα(w) α∈ n N (3.37) X (|α| + n)! = zαwα. πnα! α∈Nn We then re-index as follows: ∞ X X (|α| + n)! B(z, w) = (zw)α πnα! k=0 |α|=k ∞ X X (k + n)! = (zw)α (3.38) πnα! k=0 |α|=k ∞ X (k + 1) ··· (k + n) X k! = (zw)α. πn α! k=0 |α|=k Thus, by the multinomial theorem in conjunction with Lemma 3.1 ∞ 1 X B(z, w) = (k + 1) ··· (k + n)hz, wik πn k=0 (3.39) n! = , πn(1 − hz, wi)n+1 which is our desired result.

35 Chapter 4: The Bergman Kernel and Conformal Mappings

In this chapter we shall calculate the Bergman kernel on several other domains.

We mentioned previously that ﬁnding and explicit formula for the Bergman kernel can be a daunting task. This is mostly due to the diﬃculty in obtaining an orthonormal basis for a given domain. In some circumstances however, we may bypass this necessity altogether. The concept revolves around conformal mappings and the Riemann mapping theorem, which we will state below.

Deﬁnition 4.1. Let U, V ⊆ C. A bijective holomorphic function f : U → V is called a conformal map and we say that U and V are conformally equivalent. [3, p.

206]

The simplest examples of conformal mappings are translations (f(z) = z + a for some a ∈ C) and dilations (f(z) = cz for some c ∈ C\{0}). Take for example the function f : Dr → D deﬁned by 1 f(z) = z, (4.1) r where r > 0. Then f is a conformal mapping with inverse f −1(z) = rz.

It is useful to note that if f : U → V is a conformal map, then f 0(z) 6= 0 for all z ∈ U. Further, we also have that f −1 must also be a conformal map. They will be of particular use to us because we will show that the existence of a conformal map

36 between two domains allows you to write the Bergman kernel of one domain in terms

of the other.

Theorem 4.1. Let Ω1, Ω2 ⊆ C and suppose there exists a conformal map

f :Ω1 → Ω2. Then for all (z, w) ∈ Ω1 × Ω1,

0 0 KΩ1 (z, w) = f (z)KΩ2 (f(z), f(w))f (w), (4.2)

where KΩ1 and KΩ2 represent the Bergman kernels on Ω1 and Ω2, respectively.

2 Proof: Let {φn} be an orthonormal basis for A (Ω2). For each n ∈ N and z ∈ Ω1,

0 2 set ψn(z) = f (z)(φn ◦ f)(z). We show that {ψn} is an orthonormal basis for A (Ω1).

For if so, then we will have our result immediately from Theorem 2.16.

First let m, n ∈ N. Then Z 2 hψn, ψmiA (Ω1) = ψn(z)ψm(z) dV (z) Ω1 Z 0 0 = f (z)(φn ◦ f)(z)f (z)(φm ◦ f)(z) dV (z) (4.3) Ω1 Z 0 2 = φn(f(z))φm(f(z)) |f (z)| dV (z). Ω1 We make the change of variables w = f(z). Since f is bijective and f 0(z) 6= 0 for any

0 2 z ∈ Ω1, dV (w) = |f (z)| dV (z) and Z Z 0 2 φn(f(z))φm(f(z)) |f (z)| dV (z) = φn(w)φm(w) dV (w) Ω1 Ω2 (4.4)

2 = hφn, φmiA (Ω2).

Since {φn} is an orthonormal basis, we therefore have ( 0 m 6= n hψn, ψmi 2 = (4.5) A (Ω1) 1 m = n,

2 meaning {ψn} is an orthonormal set in A (Ω1).

37 2 2 Next suppose g ∈ A (Ω1) such that hg, ψniA (Ω1) = 0 for all n ∈ N. Then Z 0 g(z)f (z)(φn ◦ f)(z) dV (z) = 0. (4.6) Ω1

By making the same change of variables as (4.4) we have w = f(z) and dV (w) = |f 0(z)|2dV (z). Further, as f is bijective, f −1(w) = z. Thus,

Z Z −1 0 0 −1 (g ◦ f )(w) g(z)f (z)(φn ◦ f)(z) dV (z) = (f ◦ f )(w) 0 −1 2 φn(w) dV (w) Ω1 Ω2 |(f ◦ f )(w)| Z (g ◦ f −1)(w) = 0 −1 φn(w) dV (w) Ω2 (f ◦ f )(w) g ◦ f −1 = 0 −1 , φn . f ◦ f 2 A (Ω2) (4.7) g ◦ f −1 Therefore 0 −1 , φn = 0 for all n ∈ N. Since {φn} is an orthonormal f ◦ f 2 A (Ω2) g ◦ f −1 basis for A2(Ω ), we must have that = 0. Thus, (g◦f −1)(w) = g(f −1(w)) = 0 2 f 0 ◦ f −1 for every w ∈ Ω2. Therefore, as f is bijective, g(z) = 0 for all z ∈ Ω1. That is, g = 0.

As g was arbitrary under this assumption we must have that {ψn} is an orthonormal

2 basis for A (Ω1).

By Theorem 2.16,

∞ X KΩ1 (z, w) = ψn(z)ψn(w) n=0 ∞ X 0 0 = f (z)(φn ◦ f)(z)(φn ◦ f)(w) f (w) (4.8) n=0 ∞ ! 0 X 0 = f (z) φn(f(z))φn(f(w)) f (w). n=0

∞ 2 X However, as {φn} is an orthonormal basis for A (Ω2), KΩ1 (z, w) = φn(z)φn(w) by n=0 Theorem 2.16 once again. Hence,

0 0 (4.9) KΩ1 (z, w) = f (z)KΩ2 (f(z), f(w))f (w).

38 For a simple example of Theorem 4.1 we recall f : Dr → D deﬁned (4.1). Then 1 f 0(z) = and we therefore have, for all z, w ∈ , r Dr

0 0 KDr (z, w) = f (z)KD(f(z), f(w))f (w) ! 1 1 1 = r zw 2 r (4.10) π 1 − r2 r2 = . π(r2 − zw)2 Thus in some occasions we may forgo ﬁnding an orthonormal basis to compute

the Bergman kernel. Indeed, if we have an explicit formula for the Bergman kernel

of Ω2 and a conformal mapping f :Ω1 → Ω2, then we immediately have an explicit

formula for the Bergman kernel of Ω1. Of course ﬁnding a conformal mapping can be

diﬃcult, but it at least opens up another method of calculating the Bergman kernel.

Considering that we know the formula for the Bergman kernel on D, Theorem 4.1 becomes even more valuable when we consider the Riemann mapping theorem.

Theorem 4.2 (The Riemann mapping theorem). Suppose Ω ⊆ C is proper and

simply connected. If z0 ∈ Ω, then there exists a unique conformal map F :Ω → D

0 such that F (z0) = 0 and F (z0) > 0. [3, p. 224]

Thus if we know Ω ⊆ C is proper and simply connected, then there must exists a

conformal map between Ω and D. So calculating the Bergman kernel on Ω amounts

to ﬁnding a conformal map between Ω and D. In many cases the conformal maps are known explicitly.

Corollary 4.3 (The Bergman kernel for the upper half plane). The Bergman kernel

on the upper half plane H = {z ∈ C | Im(z) > 0} is given by the equation

−1 K (z, w) = (4.11) H π(z − w)2

39 for z, w ∈ H. i − z Proof: Notice that the function f : → deﬁned by f(z) = is a conformal H D i + z 1 − z map with inverse f −1(z) = i . For a proof of this, we suggest [3] on pg. 208. 1 + z −2i 2i Then f 0(z) = and f 0(z) = . Thus, by Theorem 4.1, (i + z)2 (−i + z)2

0 0 KH(z, w) = f (z)KD(f(z), f(w))f (w) ! −2i 1 2i (4.12) = . 2 i−z −i−w 2 2 (i + z) π(1 − i+z −i+w ) (−i + w) Carrying out the simpliﬁcation we see   −2i 1 2i   KH(z, w) = 2 (i + z)2  (i+z)(−i+w)−(i−z)(−i−w)  (−i + w)2 π (i+z)(−i+w)   −2i 1 2i (4.13)   = 2 (i + z)2  −2i(z−w)  (−i + w)2 π (i+z)(−i+w) 1 = . −π(z − w)2 −1 Hence, K (z, w) = . H π(z − w)2

Corollary 4.4 (The Bergman kernel for a sector). Let n ∈ N and

S = {z ∈ C | 0 < arg(z) < π/n} denote a sector in the complex plane. Then the Bergman kernel on S is given by the equation

−n2(zw)n−1 K (z, w) = (4.14) S π(zn − wn)2

for z, w ∈ S.

Proof: The function g : S → H deﬁned by g(z) = zn is a conformal map with inverse g−1(z) = z1/n. Then, by Theorem 4.1 and Corollary 4.3,

0 0 KS (z, w) = g (z)KH(g(z), g(w))g (w) −1 (4.15) = nzn−1 nwn−1. π(zn − wn)2

40 −n2(zw)n−1 Thus, K (z, w) = . S π(zn − wn)2 We ﬁnish with an example which showcases the fact the Bergman kernel is not

always given by a simple formula. In particular, we will now compute the Bergman

kernel for an arbitrary annulus (centered at the origin). Indeed, let 0 < σ < ρ < ∞

ρ ρ and consider the annulus Aσ. Note that Aσ is not simply connected and therefore we cannot utilize the Riemann mapping theorem to our advantage. Instead we must

ﬁnd an orthonormal basis.

k For each k ∈ Z deﬁne φk(z) = ckz , where q k+1  2(k+1) 2(k+1) k 6= −1 c = π(ρ −σ ) (4.16) k √ 1  ρ k = −1 2πln( σ ) Note that if k < −1, then k+1 < 0. However, as σ < ρ, we will have ρ2(k+1)−σ2(k+1)< 0

when k < −1. Thus ck is well deﬁned for k < −1.

2 ρ Theorem 4.5. The set {φk}k∈Z deﬁned above forms an orthonormal basis for A (Aσ).

Proof: We ﬁrst show that {φk}k∈Z forms an orthonormal set. Indeed, for distinct

k, l ∈ Z, Z k l hφk, φli = ckclz z dV (z). (4.17) ρ Aσ Setting z = reiθ with σ ≤ r ≤ ρ and 0 ≤ θ ≤ 2π yields Z Z ρ Z 2π k l k+l+1 i(k−l)θ ckclz z dV (z) = ckclr e dθdr. (4.18) ρ Aσ σ 0

However, since k 6= l, we have from (2.35) that hφk, φli = 0. Next we show normality.

We have two cases to consider. First suppose k 6= −1. Then Z ρ Z 2π 2 k + 1 2k+1 ||φk|| 2 ρ = r dθdr L (Aσ) 2(k+1) 2(k+1) σ 0 π(ρ − σ ) Z ρ 2(k + 1) 2k+1 (4.19) = 2(k+1) 2(k+1) r dr ρ − σ σ = 1.

41 Second, suppose k = −1. Then Z ρ Z 2π 2 1 −1 ||φ−1|| 2 ρ = r dr L (Aσ) ρ σ 0 2πln( σ ) Z ρ 1 −1 (4.20) = ρ r dr ln( σ ) σ = 1.

2 ρ Thus, ||φk||L (Aσ) = 1 for all k ∈ Z, thereby proving the orthonormality of {φk}k∈Z.

2 ρ Next we prove that {φk}k∈Z forms a basis for A (Aσ), which will require the use of Laurent series.

2 ρ 2 ρ Suppose f ∈ A (Aσ) such that hf, φki = 0 for all k ∈ Z. Then, as f ∈ A (Aσ), ∞ ρ X m f ∈ O(Aσ). Thus f has a Laurent series expansion, f(z) = amz , which converges −∞ ρ uniformly within Aσ. Fix k ∈ Z and let > 0 such that σ + < ρ − . Then, by converting to polar, Z Z ∞ ! X m k f(z)φk(z) dV (z) = amz ckz dV (z) ρ− ρ− A A σ+ σ+ −∞ (4.21) Z ρ− Z 2π ∞ ! X m imθ k −ikθ = amr e ckr e rdθdr. σ+ 0 −∞ ρ− Since f will converge uniformly on Aσ+ to its Laurent series, we have Z ρ−Z 2π ∞ ! Z ρ− ∞ Z 2π X m imθ k −ikθ X m+k+1 i(m−k)θ amr e ckr e rdθdr = amckr e dθdr. σ+ 0 −∞ σ+ −∞ 0 (4.22)

By (2.35) all terms vanish save for when m = k and we are left with Z ρ− Z 2π Z ρ− 2k+1 2k+1 akckr dθdr = 2πakck r dr σ+ 0 σ+ (4.23) πa c = k k (ρ − )2(k+1) − (σ + )2(k+1) . k + 1 ρ − Note that in the event that k = −1, we will instead have 2πa c ln( ). In any k k σ + case, as > 0 was arbitrary, we may take → 0 to have

( πakck 2(k+1) 2(k+1) k+1 ρ − σ k 6= −1 2 ρ hf, φkiA (Aσ) = ρ (4.24) 2πakckln( σ ) k = −1. 42 2 ρ By assumption hf, φkiA (Aσ) = 0, which implies ak = 0. As k ∈ Z was arbitrary, it follows that ak = 0 for all k ∈ Z. Thus f = 0 and {φk}k∈Z is an orthonormal basis

2 ρ for A (Aσ).

Therefore, by Theorem 2.16,

X 2 k k K ρ (z, w) = |c | z w , Aσ k (4.25) k∈Z

ρ where z, w ∈ Aσ and the ck are as deﬁned in (4.16). By looking at this summation it is very diﬃcult to determine whether or not this can simplify to “nice” equation such

as on D or any of the other domains we have computed previously. While it may not be as elegant as the unit disk, there is a way to represent the Bergman kernel for the

annulus in a clever way. The ﬁrst step will utilize conformal mappings.

ρ ρ Given any annulus Aσ, there exists τ > 0 such that Aσ is conformally equivalent ρ to τ . Indeed, set τ = and deﬁne f : τ → ρ by f(z) = σz. Then f is a A1 σ A1 Aσ conformal mapping between the two annuli. Thus given any annulus, we may assume

without loss of generality that it has inner radius 1.

Therefore our constants ck reduce to

q k+1  π(ρ2(k+1)−1) k 6= −1 ck = (4.26) √ 1 k = −1,  2πln(ρ)

which will simplify calculations slightly. Next we break up our summation into parts.

Namely,

−2 ∞ X k + 1 k 1 X k + 1 k K ρ (z, w) = (zw) + + (zw) . A1 π(ρ2(k+1) − 1) 2πln(ρ)zw π(ρ2(k+1) − 1) k=−∞ k=0 (4.27)

43 k + 1 For the ﬁrst summation, we add and subtract to obtain π

−2 −2 X k + 1 X k + 1 k + 1 k + 1 (zw)k = + − (zw)k π(ρ2(k+1) − 1) π(ρ2(k+1) − 1) π π k=−∞ k=−∞ (4.28) −2 −2 X (k + 1)ρ2(k+1) X k + 1 = (zw)k − (zw)k. π(ρ2(k+1) − 1) π k=−∞ k=−∞

Notice that

−2 −2 −k ∞ k X k + 1 X −k − 1 1 X k − 1 1 − (zw)k = = . (4.29) π π zw π zw k=−∞ k=−∞ k=2 1 Then, as z, w ∈ ρ, |z|, |w| > 1. Thus < 1 and we have by Lemma 2.20 that A1 |zw|

∞ k ∞ k+2 X k − 1 1 X k + 1 1 = π zw π zw k=2 k=0 ∞ k 1 X 1 = (k + 1) (4.30) π(zw)2 zw k=0 1 1 = π(zw)2 1 2 1 − zw To make a small computational aside, we note that

1 1 1 = , (4.31) π(zw)2 1 2 π(1 − zw)2 1 − zw which is the same formula for the Bergman kernel on the unit disk only now applied 1 1 to z, w ∈ ρ. In any case, since z, w ∈ ρ, , ∈ . Thus we have from (4.30) that A1 A1 z w D

−2 X k + 1 1 1 1 − (zw)k = K , (4.32) π (zw)2 D z w k=−∞ and our ﬁrst summation in (4.27) is equivalent to

−2 1 1 1 X (k + 1)ρ2(k+1) K , + (zw)k. (4.33) (zw)2 D z w π(ρ2(k+1) − 1) k=−∞

44 k + 1 For the second summation in (4.27), we add and subtract to obtain πρ2(k+1)

∞ ∞ X k + 1 X k + 1 k + 1 k + 1 (zw)k = − + (zw)k π(ρ2(k+1) − 1) π(ρ2(k+1) − 1) πρ2(k+1) πρ2(k+1) k=0 k=0 ∞ ∞ X k + 1 X k + 1 = (zw)k + (zw)k. πρ2(k+1)(ρ2(k+1) − 1) πρ2(k+1) k=0 k=0 (4.34)

However, note that

∞ ∞ k! X k + 1 1 X k + 1 zw 1 (zw)k = . (4.35) πρ2(k+1) ρ π ρ2 ρ k=0 k=0

ρ zw As z, w ∈ A1, < 1. In addition, we also have that z, w ∈ Dr. Recalling the ρ2

Bergman kernel on Dr mentioned in (4.10), we utilize Lemma 2.20 again to see that   ∞ ! 1 k + 1 zwk 1 1 1 1 X   2 = 2 ρ π ρ ρ ρ  zw  ρ (4.36) k=0 π 1 − ρ2

= KDρ (z, w). Thus our second summation in (4.27) is equivalent to

∞ X k + 1 K (z, w) + (zw)k. (4.37) Dρ πρ2(k+1)(ρ2(k+1) − 1) k=0

ρ To bring everything together we now have that for all z, w ∈ A1

1 1 1 K ρ (z, w) = K , + K (z, w) + R(z, w), (4.38) A1 (zw)2 D z w Dρ where

−2 ∞ 1 X (k + 1)ρ2(k+1) X k + 1 R(z, w) = + (zw)k + (zw)k. 2πln(ρ)zw π(ρ2(k+1) − 1) πρ2(k+1)(ρ2(k+1) − 1) k=−∞ k=0 (4.39)

Unfortunately, the series in R cannot be summed explicitly. However, R(z, w) has a few important properties. For example, each of the three pieces in the sum is

45 ρ bounded. Therefore, there exists C > 0 such that |R(z, w)| ≤ C for all z, w ∈ A1. So R can be thought of as a “remainder” of sorts. To the point, this implies that the

ρ Bergman kernel on the annulus A1 is directly related to the Bergman kernels on the unit disk and disk of radius ρ.

46 Bibliography

[1] Theodore W. Gamelin. Complex Analysis. Springer, 2001.

[2] Jaap Korevaar and Jan Wiegerinck. Several Complex Variables, 2011 (accessed April 5, 2016). https://staff.fnwi.uva.nl/j.j.o.o.wiegerinck/edu/scv/ scv1.pdf.

[3] Elias M. Stein and Rami Shakarchi. Complex Analysis. Princeton University Press, 2003.

[4] Elias M. Stein and Rami Shakarchi. Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press, 2005.